2. February 5, 2008  Go over Charles’s Law and Avogadro’s Law Homework  Introduce Combined Gas Law  Introduce Ideal Gas Law  Work Sample Problems  HOMEWORK: Pg. 480 -- #21, 23ac, 24ac, 25, 28, 29

3. Equation of State of an Ideal Gas Robert Boyle ( 1662 ) found that at fixed temperature –Pressure and volume of a gas is inversely proportional PV = constantBoyle’s Law J. Charles found that at fixed pressure –Volume of gas is proportional to change in temperature Volume Temp-273.15 o C All gases extrapolate to zero volume at a temperature corresponding to –273.15 o C (absolute zero). He CH 4 H2OH2O H2H2

4. Kelvin Temperature Scale Kelvin temperature (K) is given by K = o C + 273.15 where K is the temperature in Kelvins, o C is temperature in Celsius Using the ABSOLUTE scale, it is now possible to write Charles’ Law as V / T = constant Charles’ Law Combining Boyle’s law, Charles’ law, and another law called Gay-Lussac’s Law (relating pressure and temperature) we can mathematically prove that P V / T = constant Charles

5. Combined Gas Law This brings us to the combined gas law:

6. Practice Problem A 1.50 L sample of neon gas at 1.10 atm and 25 °C is heated to 45 °C. The neon gas is the subjected to a pressure of 1.50 atm. Determine the new volume of the neon gas. P 1 = 1.10 atm V 1 = 1.50 L T 1 = 25 °C = 298 K P 2 = 1.50 atm V 1 = ???? L T 1 = 45 °C = 318 K V 2 = 1.17 L

7. The Combined Gas Law When measured at STP, a quantity of gas has a volume of 500 cm 3. What volume will it occupy at 0 o C and 93.3 kPa? P 1 = 101.3 kPa T 1 = 273 K V 1 = 500 cm 3 P 2 = 93.3 kPa T 2 = 0 o C + 273 = 273 K V 2 = X cm 3 (101.3 kPa) x (500 cm 3 ) = (93.3 kPa) x (V 2 ) 273 K V 2 = 542.9 cm 3

8. Ideal vs. Real Gases No gas is ideal. Most gases behave ideally (almost) at pressures of approximately 1 atm or lower, when the temperature is approximately 0 °C or higher. When we do calculations, we will assume our gases are behaving as ideal gases

9. Ideal Gas Equation P V = n R T Universal Gas Constant Volume No. of moles Temperature Pressure R = 0.0821 atm L / mol K R = 8.314 kPa L / mol K Kelter, Carr, Scott, Chemistry A Wolrd of Choices 1999, page 366

10. Ideal Gas Law What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300 o C and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine T = 300 o C P = 740 mm Hg R = 0.0821 atm. L / mol. K Step 2) Equation: V= nRT P V (500 g)(0.0821 atm. L / mol. K)(300 o C) 740 mm Hg = Step 3) Solve for variable Step 4) Substitute in numbers and solve V = What MISTAKES did we make in this problem? PV = nRT

11. What mistakes did we make in this problem? What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300 o C and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine  Convert mass to moles; recall iodine is diatomic (I 2 ) 500 g I(1 mole I 2 /254 g I 2 ) n = 1.9685 mol I 2 T = 300 o C  Temperature must be converted to Kelvin T = 300 o C + 273 T = 573 K P = 740 mm Hg  Pressure needs to have same unit as R; therefore, convert pressure from mm Hg to atm. x atm = 740 mm Hg (1 atm / 760 mm Hg) P = 0.8 atm R = 0.0821 atm. L / mol. K

12. Ideal Gas Law What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300 o C and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine n = 1.9685 mol I 2 T = 573 K (300 o C) P = 0.9737 atm (740 mm Hg) R = 0.0821 atm. L / mol. K V = ? L Step 2) Equation: PV = nRT V= nRT P V (1.9685 mol)(0.0821 atm. L / mol. K)(573 K) 0.9737 atm = Step 3) Solve for variable Step 4) Substitute in numbers and solve V = 95.1 L I 2